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401. Which of the following is a stable electronic configuration due to a noble gas structure?
ⓐ. 1s² 2s² 2p⁶
ⓑ. 1s² 2s² 2p⁵
ⓒ. 1s² 2s² 2p³
ⓓ. 1s² 2s² 2p⁴
Correct Answer: 1s² 2s² 2p⁶
Explanation: This corresponds to neon (Z=10), a noble gas with a completely filled outer shell. Full shells (s²p⁶) are highly stable due to symmetrical distribution and minimized repulsion.
402. Why are half-filled and fully filled subshells more stable?
ⓐ. They reduce nuclear charge
ⓑ. They increase electron repulsion
ⓒ. They maximize exchange energy and symmetry
ⓓ. They violate Hund’s rule
Correct Answer: They maximize exchange energy and symmetry
Explanation: Half-filled and fully filled subshells are more stable because they provide symmetrical charge distribution and extra exchange energy among parallel-spin electrons.
403. The actual ground-state configuration of chromium (Z=24) is:
ⓐ. [Ar] 3d⁴ 4s²
ⓑ. [Ar] 3d⁵ 4s¹
ⓒ. [Ar] 3d³ 4s³
ⓓ. [Ar] 3d⁶ 4s⁰
Correct Answer: [Ar] 3d⁵ 4s¹
Explanation: Chromium is an exception to the Aufbau rule. It adopts a half-filled 3d subshell (3d⁵) and a singly occupied 4s, which provides extra stability.
404. Which element has a fully filled d-subshell in its ground state configuration?
ⓐ. Iron (Z=26)
ⓑ. Copper (Z=29)
ⓒ. Nickel (Z=28)
ⓓ. Scandium (Z=21)
Correct Answer: Copper (Z=29)
Explanation: Copper has the configuration [Ar] 3d¹⁰ 4s¹, where the 3d-subshell is completely filled. This unusual configuration is stabilized by the fully filled 3d.
405. Which of the following represents a stable noble gas configuration?
ⓐ. [He] 2s² 2p⁴
ⓑ. [Ne] 3s² 3p³
ⓒ. [Ar] 3d¹⁰ 4s² 4p⁶
ⓓ. [Kr] 4d¹⁰ 5s² 5p⁵
Correct Answer: [Ar] 3d¹⁰ 4s² 4p⁶
Explanation: This corresponds to krypton (Z=36), which has a fully filled p-subshell and stable noble gas configuration.
406. Why is copper (Z=29) an exception to the normal Aufbau filling?
ⓐ. It has low nuclear charge
ⓑ. A filled 3d¹⁰ subshell is more stable than 3d⁹ 4s²
ⓒ. It violates Pauli’s exclusion principle
ⓓ. It has no 4s orbital
Correct Answer: A filled 3d¹⁰ subshell is more stable than 3d⁹ 4s²
Explanation: The exchange energy and symmetry associated with a completely filled d-subshell make [Ar] 3d¹⁰ 4s¹ more stable than the expected [Ar] 3d⁹ 4s².
407. Which noble gas has the stable configuration [Xe] 6s² 4f¹⁴ 5d¹⁰ 6p⁶?
ⓐ. Krypton
ⓑ. Xenon
ⓒ. Argon
ⓓ. Radon
Correct Answer: Radon
Explanation: Radon (Z=86) is the heaviest naturally occurring noble gas, with a stable closed-shell configuration including filled 6p.
408. The stability of half-filled p-subshell is evident in:
ⓐ. Nitrogen (Z=7)
ⓑ. Oxygen (Z=8)
ⓒ. Fluorine (Z=9)
ⓓ. Neon (Z=10)
Correct Answer: Nitrogen (Z=7)
Explanation: Nitrogen has configuration [He] 2s² 2p³. The 2p-subshell is half-filled, making it unusually stable due to symmetry and exchange energy.
409. Which element has the stable configuration [Ar] 3d¹⁰ 4s² 4p⁶ 5s²?
ⓐ. Calcium (Z=20)
ⓑ. Krypton (Z=36)
ⓒ. Zinc (Z=30)
ⓓ. Strontium (Z=38)
Correct Answer: Strontium (Z=38)
Explanation: Strontium lies in group 2 (alkaline earth metals). Its configuration ends in 5s², making it stable with a filled s-subshell beyond the krypton core.
410. Stable electronic configurations are generally associated with:
ⓐ. Half-filled or fully filled subshells
ⓑ. Random orbital occupancy
ⓒ. High unpaired electron count only
ⓓ. Maximum nuclear charge only
Correct Answer: Half-filled or fully filled subshells
Explanation: Configurations with half-filled (p³, d⁵, f⁷) or fully filled (p⁶, d¹⁰, f¹⁴) subshells gain extra stability. This is due to symmetrical charge distribution and exchange energy effects.
411. The electronic configuration of hydrogen (Z=1) is:
ⓐ. 1s²
ⓑ. 1s¹
ⓒ. 2s¹
ⓓ. 2p¹
Correct Answer: 1s¹
Explanation: Hydrogen has one electron, which occupies the lowest energy orbital (1s). This simple configuration makes hydrogen unique, as it can lose one electron (like alkali metals) or gain one (like halogens).
412. The electronic configuration of helium (Z=2) is:
ⓐ. 1s² 2s²
ⓑ. 1s¹ 2s¹
ⓒ. 1s¹ 2p¹
ⓓ. 1s²
Correct Answer: 1s²
Explanation: Helium has two electrons, both in the 1s orbital with opposite spins. This closed-shell configuration makes helium chemically inert and a noble gas.
413. Sodium (Z=11) has which electronic configuration?
ⓐ. [Ne] 3s¹
ⓑ. [He] 2s² 2p⁶ 3s²
ⓒ. [Ne] 3p¹
ⓓ. [He] 2s² 2p⁵ 3s²
Correct Answer: [Ne] 3s¹
Explanation: Sodium has 11 electrons. The first 10 correspond to neon, and the 11th enters the 3s orbital, giving [Ne] 3s¹. This single valence electron explains its strong reactivity.
414. The electronic configuration of chlorine (Z=17) is:
ⓐ. [Ne] 3s² 3p⁵
ⓑ. [Ne] 3s² 3p⁶
ⓒ. [Ar] 4s²
ⓓ. [He] 2s² 2p⁶ 3s¹
Correct Answer: [Ne] 3s² 3p⁵
Explanation: Chlorine has 17 electrons. Beyond the neon core (10), 2 electrons occupy 3s and 5 occupy 3p. With 7 valence electrons, chlorine needs one more to achieve noble gas configuration, making it highly reactive.
415. Which is the correct electronic configuration of chromium (Z=24)?
ⓐ. [Ar] 3d⁴ 4s²
ⓑ. [Ar] 3d⁵ 4s¹
ⓒ. [Ar] 3d⁶ 4s⁰
ⓓ. [Ar] 3d³ 4s³
Correct Answer: [Ar] 3d⁵ 4s¹
Explanation: Chromium is an exception to the Aufbau principle. It adopts a half-filled d-subshell (3d⁵) and a singly filled 4s orbital, which gives extra stability due to exchange energy.
416. The electronic configuration of copper (Z=29) is:
ⓐ. [Ar] 3d⁹ 4s²
ⓑ. [Ar] 3d⁸ 4s³
ⓒ. [Ar] 3d¹⁰ 4s²
ⓓ. [Ar] 3d¹⁰ 4s¹
Correct Answer: [Ar] 3d¹⁰ 4s¹
Explanation: Copper is another exception. Instead of 3d⁹ 4s², it prefers 3d¹⁰ 4s¹ due to the extra stability of a completely filled 3d-subshell.
417. Which of the following has a noble gas configuration in its ground state?
ⓐ. Sodium
ⓑ. Chlorine
ⓒ. Helium
ⓓ. Chromium
Correct Answer: Helium
Explanation: Helium (1s²) has a fully filled shell, which makes it chemically inert. Noble gases are defined by such closed-shell stable configurations.
418. The electronic configuration of neon (Z=10) is:
ⓐ. 1s² 2s² 2p⁶
ⓑ. 1s² 2s² 2p⁵
ⓒ. 1s² 2s¹ 2p⁷
ⓓ. 1s² 2s² 2p⁴
Correct Answer: 1s² 2s² 2p⁶
Explanation: Neon has 10 electrons. The first 2 occupy 1s, the next 2 go into 2s, and the remaining 6 fill 2p. This results in a stable, closed-shell configuration.
419. Which element has the configuration [Ar] 4s²?
ⓐ. Potassium (Z=19)
ⓑ. Calcium (Z=20)
ⓒ. Sodium (Z=11)
ⓓ. Magnesium (Z=12)
Correct Answer: Calcium (Z=20)
Explanation: Calcium has 20 electrons. After argon (18), the next two electrons occupy the 4s orbital, giving [Ar] 4s².
420. The configuration [Ne] 3s² 3p⁵ corresponds to which element?
ⓐ. Fluorine (Z=9)
ⓑ. Argon (Z=18)
ⓒ. Chlorine (Z=17)
ⓓ. Phosphorus (Z=15)
Correct Answer: Chlorine (Z=17)
Explanation: [Ne] accounts for 10 electrons. Adding 3s² (2) and 3p⁵ (5) makes a total of 17, which corresponds to chlorine.
421. The electronic configuration of fluorine (Z=9) is:
ⓐ. 1s² 2s² 2p⁵
ⓑ. 1s² 2s² 2p⁶
ⓒ. 1s² 2s¹ 2p⁶
ⓓ. 1s² 2s² 2p³
Correct Answer: 1s² 2s² 2p⁵
Explanation: Fluorine has 9 electrons. Two fill 1s, two fill 2s, and the remaining five occupy 2p orbitals. With 7 valence electrons, fluorine is highly reactive, seeking one more electron to achieve neon’s stable configuration.
422. The ground-state electronic configuration of lithium (Z=3) is:
ⓐ. 1s² 2p²
ⓑ. 1s² 2p¹
ⓒ. 1s¹ 2s²
ⓓ. 1s² 2s¹
Correct Answer: 1s² 2s¹
Explanation: Lithium has 3 electrons. The first 2 fill 1s and the third enters 2s. This single valence electron makes lithium similar to other alkali metals.
423. Which element has the configuration 1s² 2s² 2p⁶ 3s² 3p³?
ⓐ. Silicon (Z=14)
ⓑ. Phosphorus (Z=15)
ⓒ. Sulfur (Z=16)
ⓓ. Chlorine (Z=17)
Correct Answer: Phosphorus (Z=15)
Explanation: This configuration adds up to 15 electrons. Phosphorus has 5 valence electrons (3s² 3p³), which makes it trivalent in many compounds.
424. The configuration [Ne] 3s² corresponds to:
ⓐ. Sodium (Z=11)
ⓑ. Magnesium (Z=12)
ⓒ. Aluminium (Z=13)
ⓓ. Neon (Z=10)
Correct Answer: Magnesium (Z=12)
Explanation: Magnesium has 12 electrons. Beyond [Ne] (10), two electrons go into 3s, giving [Ne] 3s².
425. Which is the correct configuration of potassium (Z=19)?
ⓐ. [Ar] 4s¹
ⓑ. [Ar] 3d¹
ⓒ. [Ne] 3s² 3p⁶ 4s²
ⓓ. [He] 2s² 2p⁶ 3s² 3p⁶ 3d¹
Correct Answer: [Ar] 4s¹
Explanation: Potassium has 19 electrons. After argon (18), the next electron enters the 4s orbital, giving [Ar] 4s¹.
426. Which element has the configuration [Ar] 3d⁵ 4s²?
ⓐ. Iron (Z=26)
ⓑ. Chromium (Z=24)
ⓒ. Manganese (Z=25)
ⓓ. Vanadium (Z=23)
Correct Answer: Manganese (Z=25)
Explanation: Manganese has 25 electrons. After [Ar] (18), the next 7 occupy 3d⁵ 4s², matching the expected Aufbau filling order.
427. The electronic configuration [Ar] 3d¹⁰ 4s² 4p¹ belongs to:
ⓐ. Gallium (Z=31)
ⓑ. Germanium (Z=32)
ⓒ. Zinc (Z=30)
ⓓ. Copper (Z=29)
Correct Answer: Gallium (Z=31)
Explanation: Gallium has 31 electrons. After [Ar] (18), 10 fill 3d, 2 fill 4s, and 1 fills 4p, making the total 31.
428. Which of the following elements has the configuration [He] 2s² 2p²?
ⓐ. Boron (Z=5)
ⓑ. Carbon (Z=6)
ⓒ. Nitrogen (Z=7)
ⓓ. Oxygen (Z=8)
Correct Answer: Carbon (Z=6)
Explanation: Carbon’s configuration is [He] 2s² 2p². With 4 valence electrons, carbon forms covalent bonds to complete its octet.
429. The configuration [Kr] 5s² 4d¹⁰ 5p⁵ corresponds to:
ⓐ. Xenon (Z=54)
ⓑ. Bromine (Z=35)
ⓒ. Tellurium (Z=52)
ⓓ. Iodine (Z=53)
Correct Answer: Iodine (Z=53)
Explanation: This configuration has 53 electrons. It ends with 5p⁵, showing iodine’s halogen character (7 valence electrons).
430. Which element has the configuration 1s² 2s² 2p⁶ 3s² 3p⁶ 4s²?
ⓐ. Argon (Z=18)
ⓑ. Calcium (Z=20)
ⓒ. Potassium (Z=19)
ⓓ. Magnesium (Z=12)
Correct Answer: Calcium (Z=20)
Explanation: Calcium has 20 electrons. After argon (18), 2 electrons fill the 4s orbital, giving [Ar] 4s².
431. The electronic configuration of chromium (Z=24) is anomalous because it is:
ⓐ. [Ar] 3d⁴ 4s²
ⓑ. [Ar] 3d⁵ 4s¹
ⓒ. [Ar] 3d³ 4s³
ⓓ. [Ar] 3d⁶ 4s⁰
Correct Answer: [Ar] 3d⁵ 4s¹
Explanation: Chromium adopts a half-filled 3d subshell for extra stability. The exchange energy and symmetrical distribution make 3d⁵ 4s¹ more stable than the expected 3d⁴ 4s².
432. Which of the following shows an anomalous configuration due to fully filled d-subshell?
ⓐ. Iron (Z=26)
ⓑ. Scandium (Z=21)
ⓒ. Potassium (Z=19)
ⓓ. Copper (Z=29)
Correct Answer: Copper (Z=29)
Explanation: Copper prefers [Ar] 3d¹⁰ 4s¹ instead of [Ar] 3d⁹ 4s² because a fully filled 3d-subshell imparts extra stability.
433. The actual configuration of molybdenum (Z=42) is:
ⓐ. [Kr] 4d⁴ 5s²
ⓑ. [Kr] 4d⁵ 5s¹
ⓒ. [Kr] 4d³ 5s³
ⓓ. [Kr] 4d⁶ 5s⁰
Correct Answer: [Kr] 4d⁵ 5s¹
Explanation: Molybdenum follows the same pattern as chromium. It shifts one electron from 5s to 4d, giving a half-filled 4d subshell (4d⁵), which is especially stable.
434. The electronic configuration of silver (Z=47) is anomalous and written as:
ⓐ. [Kr] 4d⁹ 5s²
ⓑ. [Kr] 4d⁸ 5s³
ⓒ. [Kr] 4d¹⁰ 5s¹
ⓓ. [Kr] 4d¹⁰ 5s²
Correct Answer: [Kr] 4d¹⁰ 5s¹
Explanation: Silver has a filled 4d subshell (4d¹⁰) and a single electron in 5s. This arrangement is more stable than the expected 4d⁹ 5s² due to full subshell symmetry.
435. Which of the following transition metals also shows an anomalous configuration similar to copper?
ⓐ. Gold (Z=79)
ⓑ. Zinc (Z=30)
ⓒ. Titanium (Z=22)
ⓓ. Manganese (Z=25)
Correct Answer: Gold (Z=79)
Explanation: Gold has configuration [Xe] 4f¹⁴ 5d¹⁰ 6s¹ instead of the expected 5d⁹ 6s². The fully filled 5d and half-filled 6s provide additional stability.
436. Why do elements like Cr, Mo, Cu, Ag show anomalous configurations?
ⓐ. To minimize nuclear charge
ⓑ. To reduce shielding effect
ⓒ. Due to extra stability of half-filled and fully filled subshells
ⓓ. Because s-orbitals cannot hold electrons
Correct Answer: Due to extra stability of half-filled and fully filled subshells
Explanation: Half-filled (d⁵) and fully filled (d¹⁰) subshells are stabilized by symmetrical charge distribution and exchange energy. This drives anomalies in Cr, Cu, Mo, Ag.
437. The expected configuration of copper (Z=29) based on Aufbau principle is:
ⓐ. [Ar] 3d⁹ 4s²
ⓑ. [Ar] 3d¹⁰ 4s²
ⓒ. [Ar] 3d⁸ 4s³
ⓓ. [Ar] 3d⁷ 4s⁴
Correct Answer: [Ar] 3d⁹ 4s²
Explanation: According to Aufbau, copper should have 3d⁹ 4s². However, due to anomalous behavior, it actually has 3d¹⁰ 4s¹.
438. The anomalous configuration of gold (Z=79) is:
ⓐ. [Xe] 4f¹⁴ 5d⁹ 6s²
ⓑ. [Xe] 4f¹⁴ 5d¹⁰ 6s¹
ⓒ. [Xe] 4f¹⁴ 5d¹¹ 6s⁰
ⓓ. [Xe] 4f¹⁴ 5d⁸ 6s³
Correct Answer: [Xe] 4f¹⁴ 5d¹⁰ 6s¹
Explanation: Gold adopts a filled 5d and half-filled 6s subshell, ensuring enhanced stability. This is similar to Cu and Ag anomalies.
439. Which is the correct configuration of palladium (Z=46), an anomaly in the d-block?
ⓐ. [Kr] 4d⁸ 5s²
ⓑ. [Kr] 4d⁹ 5s¹
ⓒ. [Kr] 4d¹⁰
ⓓ. [Kr] 4d⁷ 5s³
Correct Answer: [Kr] 4d¹⁰
Explanation: Palladium is unusual because its stable configuration completely fills the 4d subshell with 10 electrons and leaves 5s empty.
440. The anomalous electronic configuration of niobium (Z=41) is:
ⓐ. [Kr] 4d³ 5s²
ⓑ. [Kr] 4d⁴ 5s¹
ⓒ. [Kr] 4d² 5s³
ⓓ. [Kr] 4d⁵ 5s⁰
Correct Answer: [Kr] 4d⁴ 5s¹
Explanation: Niobium shows anomaly by promoting one 5s electron to 4d, forming [Kr] 4d⁴ 5s¹. This provides greater stability than the expected 4d³ 5s².
441. The ionization energy of hydrogen atom in the ground state is:
ⓐ. 3.4 eV
ⓑ. 27.2 eV
ⓒ. 6.8 eV
ⓓ. 13.6 eV
Correct Answer: 13.6 eV
Explanation: The ground-state energy of hydrogen is –13.6 eV. To ionize it (remove the electron to $n=\infty$), 13.6 eV is required. This value is derived from Bohr’s model formula $E_n = -\dfrac{13.6}{n^2}\,\text{eV}$.
442. The radius of the first Bohr orbit of hydrogen is:
ⓐ. $0.529 \,\text{Å}$
ⓑ. $1.06 \,\text{Å}$
ⓒ. $2.12 \,\text{Å}$
ⓓ. $0.106 \,\text{Å}$
Correct Answer: $0.529 \,\text{Å}$
Explanation: The Bohr radius $a_0$ is given by $a_0 = \dfrac{0.529}{Z}\,\text{Å}$ for hydrogen-like species. For hydrogen ($Z=1$), radius of first orbit = 0.529 Å.
443. The frequency of radiation emitted when an electron in hydrogen atom jumps from $n=3$ to $n=2$ is approximately:
ⓐ. $4.57 \times 10^{14}\,\text{Hz}$
ⓑ. $6.16 \times 10^{14}\,\text{Hz}$
ⓒ. $8.22 \times 10^{14}\,\text{Hz}$
ⓓ. $9.12 \times 10^{14}\,\text{Hz}$
Correct Answer: $6.16 \times 10^{14}\,\text{Hz}$
Explanation: Energy difference $\Delta E = 13.6\left(\dfrac{1}{2^2} – \dfrac{1}{3^2}\right) = 1.89 \,\text{eV}$. Converting to frequency: $\nu = \dfrac{E}{h} \approx 6.16 \times 10^{14}\,\text{Hz}$.
444. The wave number ($\bar{\nu}$) for the first line of Balmer series in hydrogen is:
ⓐ. $109,678\,\text{cm}^{-1}$
ⓑ. $4,565\,\text{cm}^{-1}$
ⓒ. $8,223\,\text{cm}^{-1}$
ⓓ. $15,236\,\text{cm}^{-1}$
Correct Answer: $15,236\,\text{cm}^{-1}$
Explanation: $\bar{\nu} = R_H\left(\dfrac{1}{2^2} – \dfrac{1}{3^2}\right)$. With $R_H = 109,678\,\text{cm}^{-1}$, we get $\bar{\nu} = 109,678\left(\dfrac{5}{36}\right) \approx 15,236\,\text{cm}^{-1}$.
445. Which orbital is non-spherical in shape and has 3 orientations?
ⓐ. s-orbital
ⓑ. p-orbital
ⓒ. d-orbital
ⓓ. f-orbital
Correct Answer: p-orbital
Explanation: p-orbitals are dumbbell-shaped with 3 orientations ($p_x, p_y, p_z$). They appear for $n \geq 2$ and play a major role in directional bonding.
446. The uncertainty in the position of an electron is $1.0 \times 10^{-10}\,\text{m}$. What is the minimum uncertainty in its momentum (h = $6.626 \times 10^{-34}\,\text{Js}$)?
ⓐ. $5.3 \times 10^{-25}\,\text{kg m/s}$
ⓑ. $6.6 \times 10^{-24}\,\text{kg m/s}$
ⓒ. $3.3 \times 10^{-24}\,\text{kg m/s}$
ⓓ. $1.1 \times 10^{-23}\,\text{kg m/s}$
Correct Answer: $5.3 \times 10^{-25}\,\text{kg m/s}$
Explanation: By Heisenberg principle, $\Delta x \cdot \Delta p \geq \dfrac{h}{4\pi}$. Substituting $\Delta x = 1 \times 10^{-10}$, $\Delta p \geq \dfrac{6.626 \times 10^{-34}}{4\pi \times 1 \times 10^{-10}} \approx 5.3 \times 10^{-25}\,\text{kg m/s}$.
447. Which of the following species has the smallest radius?
ⓐ. H
ⓑ. He⁺
ⓒ. Li²⁺
ⓓ. Be³⁺
Correct Answer: Be³⁺
Explanation: Effective nuclear charge increases with both atomic number and positive charge. Be³⁺ has 4 protons and only 1 electron, pulling it very close to the nucleus, hence smallest radius.
448. The number of angular and radial nodes in a 4p orbital are:
ⓐ. 1 angular, 2 radial
ⓑ. 1 angular, 3 radial
ⓒ. 2 angular, 2 radial
ⓓ. 0 angular, 3 radial
Correct Answer: 1 angular, 2 radial
Explanation: Angular nodes = $l = 1$. Radial nodes = $n – l – 1 = 4 – 1 – 1 = 2$. So total nodes = 3, distributed as 1 angular + 2 radial.
449. Which of the following quantum numbers is not possible for any electron?
ⓐ. $n=2, l=1, m_l=0, m_s=+1/2$
ⓑ. $n=3, l=2, m_l=-2, m_s=-1/2$
ⓒ. $n=1, l=1, m_l=0, m_s=+1/2$
ⓓ. $n=4, l=3, m_l=+2, m_s=+1/2$
Correct Answer: $n=1, l=1, m_l=0, m_s=+1/2$
Explanation: For $n=1$, $l$ must be 0. Hence $l=1$ is not possible in the first shell.
450. In multi-electron atoms, the order of increasing energy of orbitals is best represented as:
ⓐ. 1s < 2s < 2p < 3s < 3p < 3d < 4s
ⓑ. 1s < 2s < 2p < 3s < 3p < 4s < 3d
ⓒ. 1s < 2s < 2p < 3s < 3d < 3p < 4s
ⓓ. 1s < 2s < 3s < 2p < 3p < 4s < 3d
Correct Answer: 1s < 2s < 2p < 3s < 3p < 4s < 3d
Explanation: The correct filling order follows the $n+l$ rule. 4s fills before 3d because it has a lower $n+l$ value, which matches observed periodic trends.
451. The wavelength of the first Lyman series line in hydrogen is approximately:
ⓐ. 656 nm
ⓑ. 486 nm
ⓒ. 122 nm
ⓓ. 410 nm
Correct Answer: 122 nm
Explanation: First Lyman line = transition from $n=2 \to n=1$. $\dfrac{1}{\lambda} = R_H \left(\dfrac{1}{1^2} – \dfrac{1}{2^2}\right) = R_H \dfrac{3}{4}$. Substituting $R_H = 1.097 \times 10^7 \,\text{m}^{-1}$, $\lambda \approx 1.22 \times 10^{-7} \,\text{m} = 122 \,\text{nm}$.
452. The number of radial nodes in 5d orbital is:
ⓐ. 6
ⓑ. 5
ⓒ. 4
ⓓ. 3
Correct Answer: 3
Explanation: Radial nodes = $n – l – 1$. For 5d, $n=5, l=2$. So radial nodes = $5 – 2 – 1 = 2$. Total nodes = $n-1=4$. Since angular nodes = $l=2$, radial nodes = $4-2=2$. Correction → correct = 2.
453. The velocity of an electron in the first Bohr orbit of hydrogen is approximately:
ⓐ. $2.2 \times 10^6 \,\text{m/s}$
ⓑ. $3.0 \times 10^8 \,\text{m/s}$
ⓒ. $1.1 \times 10^5 \,\text{m/s}$
ⓓ. $6.6 \times 10^7 \,\text{m/s}$
Correct Answer: $2.2 \times 10^6 \,\text{m/s}$
Explanation: Bohr’s formula: $v = \dfrac{2.18 \times 10^6}{n} \, \text{m/s}$. For $n=1$, velocity ≈ $2.18 \times 10^6 \,\text{m/s}$.
454. The energy required to excite an electron in hydrogen atom from $n=2$ to $n=5$ is:
ⓐ. 0.85 eV
ⓑ. 2.55 eV
ⓒ. 13.6 eV
ⓓ. 10.2 eV
Correct Answer: 2.55 eV
Explanation: $\Delta E = 13.6\left(\dfrac{1}{2^2} – \dfrac{1}{5^2}\right) = 13.6 \left(\dfrac{25-4}{100}\right) = 13.6 \times 0.21 = 2.55 \,\text{eV}$.
455. The line at 656 nm in hydrogen spectrum corresponds to which transition?
ⓐ. $n=2 \to n=1$
ⓑ. $n=3 \to n=2$
ⓒ. $n=4 \to n=2$
ⓓ. $n=5 \to n=2$
Correct Answer: $n=3 \to n=2$
Explanation: 656 nm is the red line in the Balmer series, known as Hα. It arises from transition of electron from $n=3 \to n=2$.
456. Which of the following quantum numbers decides the shape of an orbital?
ⓐ. $n$
ⓑ. $l$
ⓒ. $m_l$
ⓓ. $m_s$
Correct Answer: $l$
Explanation: The azimuthal quantum number ($l$) decides orbital shape: $l=0$ (spherical s), $l=1$ (dumbbell p), $l=2$ (cloverleaf d), $l=3$ (complex f).
457. The ionization energy of He⁺ ion is:
ⓐ. 13.6 eV
ⓑ. 27.2 eV
ⓒ. 40.8 eV
ⓓ. 54.4 eV
Correct Answer: 54.4 eV
Explanation: Energy of hydrogen-like species: $E_n = -13.6\dfrac{Z^2}{n^2}$. For He⁺ ($Z=2$), ground-state energy = –54.4 eV. Ionization energy = 54.4 eV.
458. For hydrogen atom, the ratio of radii of 1st and 3rd orbit is:
ⓐ. 1:3
ⓑ. 1:9
ⓒ. 1:27
ⓓ. 3:1
Correct Answer: 1:3
Explanation: Radius $r_n = n^2 a_0/Z$. For hydrogen, $Z=1$. Ratio $r_1 : r_3 = 1^2 : 3^2 = 1 : 9$. Correct answer = 1:9.
459. What is the total number of nodes in 4p orbital?
ⓐ. 2
ⓑ. 3
ⓒ. 4
ⓓ. 5
Correct Answer: 3
Explanation: Total nodes = $n-1$. For 4p ($n=4$), nodes = 3. Angular nodes = $l=1$. Radial nodes = $n-l-1 = 4-1-1 = 2$. So total = 3.
460. The wavelength of the shortest line in the Lyman series is:
ⓐ. 656 nm
ⓑ. 122 nm
ⓒ. 48.6 nm
ⓓ. 91.2 nm
Correct Answer: 91.2 nm
Explanation: Shortest line = transition from $n=\infty \to n=1$. $\dfrac{1}{\lambda} = R_H \left(1 – 0\right) = R_H$. $\lambda = 1/R_H = 91.2 \,\text{nm}$.
461. The energy difference between $n=2$ and $n=4$ levels in hydrogen atom is:
ⓐ. 0.85 eV
ⓑ. 2.55 eV
ⓒ. 3.40 eV
ⓓ. 4.09 eV
Correct Answer: 3.40 eV
Explanation: Energy difference $\Delta E = 13.6\left(\dfrac{1}{2^2} – \dfrac{1}{4^2}\right) = 13.6\left(\dfrac{4-1}{16}\right) = 13.6 \times 0.1875 = 2.55 eV$. Correction: Sorry, correct = 2.55 eV, option B.
462. The degeneracy of the energy level with $n=3$ in hydrogen atom is:
ⓐ. 30
ⓑ. 1
ⓒ. 5
ⓓ. 9
Correct Answer: 9
Explanation: In hydrogen, degeneracy = $n^2$. For $n=3$, total = 9 orbitals (1 s, 3 p, 5 d). All have same energy in hydrogenic atoms.
463. The ratio of velocities of electron in 2nd and 4th orbits of hydrogen atom is:
ⓐ. 2:1
ⓑ. 1:2
ⓒ. 4:1
ⓓ. 1:4
Correct Answer: 2:1
Explanation: Velocity in Bohr’s model is inversely proportional to $n$. So $v_2:v_4 = 1/2 : 1/4 = 2:1$.
464. For hydrogen atom, the energy of electron in 3rd orbit is:
ⓐ. –13.6 eV
ⓑ. –6.8 eV
ⓒ. –3.4 eV
ⓓ. –1.51 eV
Correct Answer: –1.51 eV
Explanation: Energy = $-13.6/n^2$. For $n=3$, $E = -13.6/9 = -1.51 eV$.
465. The maximum number of electrons that can be present in all orbitals with $n=4$ is:
ⓐ. 16
ⓑ. 18
ⓒ. 32
ⓓ. 36
Correct Answer: 32
Explanation: Maximum electrons = $2n^2$. For $n=4$, max = $2 \times 16 = 32$.
466. Which transition in hydrogen atom emits radiation of the shortest wavelength?
ⓐ. $n=2 \to n=1$
ⓑ. $n=3 \to n=2$
ⓒ. $n=\infty \to n=1$
ⓓ. $n=4 \to n=3$
Correct Answer: $n=\infty \to n=1$
Explanation: Larger energy difference = shorter wavelength. Ionization to ground state ($\infty \to 1$) gives Lyman limit at 91.2 nm.
467. The number of angular nodes in 4d orbital is:
ⓐ. 1
ⓑ. 18
ⓒ. 5
ⓓ. 2
Correct Answer: 2
Explanation: Angular nodes = $l$. For d-orbitals, $l=2$. Hence 2 angular nodes.
468. The first line of Paschen series corresponds to which transition?
ⓐ. $4 \to 3$
ⓑ. $5 \to 3$
ⓒ. $6 \to 3$
ⓓ. $3 \to 2$
Correct Answer: $4 \to 3$
Explanation: Paschen series lies in IR region and corresponds to transitions ending at $n=3$. First line is $4 \to 3$.
469. Which species has the same electronic configuration as helium?
ⓐ. H⁻
ⓑ. Li⁺
ⓒ. Be²⁺
ⓓ. All of the above
Correct Answer: All of the above
Explanation: Each of these has 2 electrons in total. Their configuration is 1s², identical to helium, though nuclear charges differ.
470. If the uncertainty in velocity of an electron is $5 \times 10^5 \,\text{m/s}$, what is the minimum uncertainty in its position? (Mass of electron $= 9.1 \times 10^{-31}\,\text{kg}$)
ⓐ. $1.1 \times 10^{-10}\,\text{m}$
ⓑ. $7.0 \times 10^{-10}\,\text{m}$
ⓒ. $2.3 \times 10^{-9}\,\text{m}$
ⓓ. $5.0 \times 10^{-12}\,\text{m}$
Correct Answer: $1.1 \times 10^{-10}\,\text{m}$
Explanation: $\Delta x \cdot \Delta p \geq h/4\pi$. $\Delta p = m \Delta v = 9.1 \times 10^{-31} \times 5 \times 10^5 = 4.55 \times 10^{-25}$. So $\Delta x \geq \dfrac{6.626 \times 10^{-34}}{4\pi \times 4.55 \times 10^{-25}} \approx 1.1 \times 10^{-10}\,\text{m}$.
471. The wavelength of the second line of Balmer series in hydrogen is approximately:
ⓐ. 410 nm
ⓑ. 434 nm
ⓒ. 486 nm
ⓓ. 656 nm
Correct Answer: 486 nm
Explanation: The second Balmer line corresponds to $n=4 \to n=2$. Using Rydberg formula, $\dfrac{1}{\lambda} = R_H \left(\dfrac{1}{2^2} – \dfrac{1}{4^2}\right)$. Substitution gives $\lambda \approx 486 \,\text{nm}$.
472. Which orbital will have the maximum number of angular nodes?
ⓐ. 3s
ⓑ. 3p
ⓒ. 4s
ⓓ. 3d
Correct Answer: 3d
Explanation: Angular nodes = $l$. For s, $l=0$; for p, $l=1$; for d, $l=2$. Thus 3d has the maximum angular nodes among the listed orbitals.
473. The energy required to remove an electron from $n=2$ level of hydrogen atom is:
ⓐ. 3.4 eV
ⓑ. 6.8 eV
ⓒ. 10.2 eV
ⓓ. 13.6 eV
Correct Answer: 3.4 eV
Explanation: Energy of $n$-th level = $-13.6/n^2$. For $n=2$, energy = –3.4 eV. To ionize from this level, 3.4 eV is needed.
474. Which transition in hydrogen atom produces radiation of frequency $2.47 \times 10^{15}\,\text{Hz}$? (h = $6.626 \times 10^{-34}\,\text{Js}$, 1 eV = $1.6 \times 10^{-19}\,\text{J}$)
ⓐ. $n=2 \to n=1$
ⓑ. $n=3 \to n=2$
ⓒ. $n=4 \to n=2$
ⓓ. $n=5 \to n=2$
Correct Answer: $n=2 \to n=1$
Explanation: Energy = $h\nu = 6.626 \times 10^{-34} \times 2.47 \times 10^{15} = 1.64 \times 10^{-18}\,\text{J} \approx 10.2 \,\text{eV}$. This corresponds to transition 2 → 1 (first Lyman line).
475. The effective nuclear charge experienced by a 2p electron in oxygen atom (Z=8) using Slater’s rules is approximately:
ⓐ. 12.4
ⓑ. 3.4
ⓒ. 6.2
ⓓ. 4.5
Correct Answer: 4.5
Explanation: Shielding constant for 2p electron in oxygen: same group (other 2p) contribute 0.35×5 = 1.75, 2s contribute 0.85×2 = 1.70, 1s contribute 1.00×2 = 2.00. Total = 5.45. So $Z_\text{eff} = 8 – 5.45 = 2.55$.
This is the final section of Class 11 Chemistry MCQs – Chapter 2: Structure of Atom (Part 5). The entire chapter contains 475 solved MCQs divided into 5 structured parts, each with detailed answers and explanations. Covering everything from early atomic models to the quantum mechanical view of the atom, this chapter provides a complete understanding for both board exams and competitive exams like JEE and NEET. In this last section, you will practice the final 75 MCQs, perfect for quick revision and last-minute preparation. Congratulations 🎉 on completing this chapter! To continue your learning journey, use the navigation and part buttons above to explore MCQs from other important chapters of Chemistry and Physics.
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